for the polynomial, list each real zero and its multiplicity. determine whether the graph crosses or touches the x-axis at each x-intercept. f(x)=1/2x^4(x^2-5)(x+1)
f(x)=1/2^4(x^2-5)(X+1)
so what is the solution
To find the real zeros and their multiplicities of the polynomial function f(x) = 1/2x^4(x^2-5)(x+1), we need to first factorize the polynomial completely.
The given polynomial can be factored as follows:
f(x) = 1/2x^4(x^2-5)(x+1)
The factors are x^4, (x^2-5), and (x+1).
To find the real zeros, we set each factor equal to zero and solve for x:
1. x^4 = 0
In this case, x = 0 would be a real zero with multiplicity 4. The graph touches the x-axis at x = 0.
2. x^2 - 5 = 0
Rearranging the equation, we have:
x^2 = 5
Taking the square root of both sides:
x = √5 (positive square root) or x = -√5 (negative square root)
These are the real zeros with multiplicity 2. The graph crosses the x-axis at x = √5 and x = -√5.
3. x + 1 = 0
Solving for x, we get:
x = -1
This is a real zero with multiplicity 1. The graph crosses the x-axis at x = -1.
So, the real zeros and their multiplicities are:
- x = 0 (multiplicity 4) - graph touches the x-axis at x = 0
- x = √5 and x = -√5 (both with multiplicity 2) - graph crosses the x-axis at x = √5 and x = -√5
- x = -1 (multiplicity 1) - graph crosses the x-axis at x = -1